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a) (x-1)2 = 0
=> x - 1 =0 => x = 1
b) (x-5)3 = 8 = 23
=> x-5 = 2 => x = 7
c) 2x + 14 = 81
2x = 80
x = 40
d) (x-4)3 =(x-4)6
=> (x-4)6 -(x-4)3 = 0
(x-4)3. [ (x-4)3 -1 ] = 0
=> (x-4)3 = 0 => x - 4 = 0 => x = 4
(x-4)3 - 1 = 0 => (x-4)3 = 1 => x - 4 = 1 => x = 5
KL:...
Ta có :
22x - 1 : 4 = 83
=> 22x-1 : 22 = ( 23 )3
=> 22x-1 : 22 = 29
=> 22x-1 = 29 . 22 = 211
=> 2x - 1 = 11
=> 2x = 12
=> x = 6
1, \(x^2+2x-3=x^2+3x-x-3=x\left(x-1\right)+3\left(x-1\right)=\left(x+3\right)\left(x-1\right)\)
2, \(x^2+3x-10=x^2+5x-2x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)
3, \(x^2-x-12=x^2-4x+3x-12=x\left(x+3\right)-4\left(x+3\right)=\left(x-4\right)\left(x+3\right)\)
4, \(3x^2+4x-7=3x^2+7x-3x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(3x+7\right)\left(x-1\right)\)
5, \(4x^2-9y^2-5xy=4x^2-9xy+4xy-9y^2\)
\(=4x\left(x+y\right)-9y\left(x+y\right)=\left(4x-9y\right)\left(x+y\right)\)
6, \(x^2-2x-4y^2-4y=x^2-2x+1-4y^2-4y-1=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x-1-2y-1\right)\left(x-1+2y+1\right)=\left(x-2y-2\right)\left(x+2y\right)\)
2x+4=23
2x+4=8
2x =4
x =2
TL :
2x + 4 = 23
2 . x + 4 = 8
2 . x = 8 - 4
2 . x = 4
x = 4 : 2
x = 2
_HT_